\newproblem{lay:7_2_26}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 7.2.26}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Show that if an $n\times n$ matrix $A$ is positive definite, then there exists a positive definite matrix $B$ such that $A=B^TB$. [\textit{Hint}: Write $A=PDP^T$,
	with $P^T=P^{-1}$. Produce a diagonal matrix $C$ such that $D=C^TC$, and let $B=PCP^T$. Show that $B$ works.]
}{
   % Solution
	If $A$ is positive definite, then it is symmetric and it can be orthogonally diagonalized as
	\begin{center}
		$A=PDP^T$
	\end{center}
	Since it is positive definite, all its eigenvalues are larger than 0. So the diagonal matrix $D$ has all its diagonal entries larger than 0. We now define
	\begin{center}
		$C=D^{\frac{1}{2}}=\begin{pmatrix} \lambda_1^{\frac{1}{2}} & 0 & ... & 0 \\ 0 & \lambda_2^{\frac{1}{2}} & ... & 0 \\ ... & ... & ... & ... \\
		   0 & 0 & ... & \lambda_n^{\frac{1}{2}}\end{pmatrix}$
	\end{center}
	It can be easily verified that
	\begin{center}
		$C^TC=D$
	\end{center}
	We now construct
	\begin{center}
		$B=PCP^T$
	\end{center}
	Let's check that $B^TB=A$
	\begin{center}
		$B^TB=(PCP^T)^TPCP^T=(P^T)^TC^TP^TPCP^T=PC^TCP^T=PDP^T=A$
	\end{center}
}
\useproblem{lay:7_2_26}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}

